Study and Verification of Hardy-Weinberg Law by Chi-Square Analysis

 

4. Study and Verification of Hardy-Weinberg Law by Chi-Square Analysis

Example: Seed Color in Pea Plants (Mendel’s classic experiment)

In pea plants, yellow seed color (Y) is dominant over green (y). There are 3 genotypes: YY – Homozygous dominant (Yellow),  Yy – Heterozygous (Yellow), yy – Homozygous recessive (Green)

Observed Data (from an experimental population):

Phenotype	Observed Number
Yellow	72
Green	28
Total	100

check whether this population is in Hardy-Weinberg equilibrium.

Solution:

Step 1: Calculate Allele Frequencies

Let: 

Total individuals = 100

Green = yy = 28

Therefore, frequency of genotype yy (q²) = 28/100 = 0.28

From this: 

q = √0.28 = 0.529 

p = 1 – q = 1 – 0.529 = 0.471

Now, expected genotype frequencies:

• YY (p²) = (0.471)² = 0.222 → Expected count = 22.2

• Yy (2pq) = 2 × 0.471 × 0.529 = 0.498 → Expected count = 49.8

• yy (q²) = 0.28 → Expected count = 28

Step 2: Convert to Expected Phenotypes

Phenotype	Expected Number
Yellow (YY + Yy) = 22.2 + 49.8 = 72
Green (yy) = 28

Step 3: Apply Chi-square Test

Chi-square (χ²) formula:

$$χ² = \sum \frac{(O - E)^2}{E}$$

Where:

• O = Observed count

• E = Expected count

$$χ² = \frac{(72 - 72)^2}{72} + \frac{(28 - 28)^2}{28} = 0 + 0 = **0**$$

Conclusion:

χ² = 0, which is less than the critical value (3.84 at 1 d.f. and 0.05 significance level).

So, no significant difference between observed and expected data.

 The population is in Hardy-Weinberg equilibrium.